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Question
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 − 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
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Solution
According to the Arrhenius equation,
k = `Ae^(-E_a//RT)`
or, loge k = `log_e A - E_a/(RT)`
or, log10 k = `log_10 A - E_a/(2.303 RT)` ...(i)
The given equation is:
log10 k = `14.34 - (1.25 xx 10^4 (K))/T` ...(ii)
Comparing equations (i) and (ii), we have,
`E_a/(2.303 RT) = (1.25 xx 10^4 (K))/T`
or, Ea = 2.303 × R × 1.25 × 104 (K)
= 2.303 × (8.314 × 10−3 kJ K−1 mol−1) × 1.25 × 104 (K)
= 239.34 kJ mol−1
When t1/2 = 256 min, i.e., 256 × 60 s, we have
k = `0.693/t_(1//2)`
= `0.693/(256 xx 60)`
= 4.51 × 10−5 s−1
Substituting this value in equation (ii), we have,
log10 (4.51 × 10−5) = `14.34 - (1.25 xx 10^4 (K))/T`
⇒ −4.346 = `14.34 - (1.25 xx 10^4 (K))/T`
⇒ `(1.25 xx 10^4 (K))/T` = 14.34 + 4.346
⇒ `(1.25 xx 10^4 (K))/T` = 18.686
T = `(1.25 xx 10^4 (K))/18.686`
= 668.95 K
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